Welcome to Subscribe On Youtube
1905. Count Sub Islands
Description
You are given two m x n
binary matrices grid1
and grid2
containing only 0
's (representing water) and 1
's (representing land). An island is a group of 1
's connected 4directionally (horizontal or vertical). Any cells outside of the grid are considered water cells.
An island in grid2
is considered a subisland if there is an island in grid1
that contains all the cells that make up this island in grid2
.
Return the number of islands in grid2
that are considered subislands.
Example 1:
Input: grid1 = [[1,1,1,0,0],[0,1,1,1,1],[0,0,0,0,0],[1,0,0,0,0],[1,1,0,1,1]], grid2 = [[1,1,1,0,0],[0,0,1,1,1],[0,1,0,0,0],[1,0,1,1,0],[0,1,0,1,0]] Output: 3 Explanation: In the picture above, the grid on the left is grid1 and the grid on the right is grid2. The 1s colored red in grid2 are those considered to be part of a subisland. There are three subislands.
Example 2:
Input: grid1 = [[1,0,1,0,1],[1,1,1,1,1],[0,0,0,0,0],[1,1,1,1,1],[1,0,1,0,1]], grid2 = [[0,0,0,0,0],[1,1,1,1,1],[0,1,0,1,0],[0,1,0,1,0],[1,0,0,0,1]] Output: 2 Explanation: In the picture above, the grid on the left is grid1 and the grid on the right is grid2. The 1s colored red in grid2 are those considered to be part of a subisland. There are two subislands.
Constraints:
m == grid1.length == grid2.length
n == grid1[i].length == grid2[i].length
1 <= m, n <= 500
grid1[i][j]
andgrid2[i][j]
are either0
or1
.
Solutions
Solution 1: DFS
We can traverse each cell $(i, j)$ in the matrix grid2
. If the value of the cell is $1$, we start a depthfirst search from this cell, set the value of all cells connected to this cell to $0$, and record whether the corresponding cell in grid1
is also $1$ for all cells connected to this cell. If it is $1$, it means that this cell is also an island in grid1
, otherwise it is not. Finally, we count the number of subislands in grid2
.
The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the number of rows and columns of the matrices grid1
and grid2
, respectively.

class Solution { private final int[] dirs = {1, 0, 1, 0, 1}; private int[][] grid1; private int[][] grid2; private int m; private int n; public int countSubIslands(int[][] grid1, int[][] grid2) { m = grid1.length; n = grid1[0].length; this.grid1 = grid1; this.grid2 = grid2; int ans = 0; for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { if (grid2[i][j] == 1) { ans += dfs(i, j); } } } return ans; } private int dfs(int i, int j) { int ok = grid1[i][j]; grid2[i][j] = 0; for (int k = 0; k < 4; ++k) { int x = i + dirs[k], y = j + dirs[k + 1]; if (x >= 0 && x < m && y >= 0 && y < n && grid2[x][y] == 1) { ok &= dfs(x, y); } } return ok; } }

class Solution { public: int countSubIslands(vector<vector<int>>& grid1, vector<vector<int>>& grid2) { int m = grid1.size(), n = grid1[0].size(); int ans = 0; int dirs[5] = {1, 0, 1, 0, 1}; function<int(int, int)> dfs = [&](int i, int j) { int ok = grid1[i][j]; grid2[i][j] = 0; for (int k = 0; k < 4; ++k) { int x = i + dirs[k], y = j + dirs[k + 1]; if (x >= 0 && x < m && y >= 0 && y < n && grid2[x][y]) { ok &= dfs(x, y); } } return ok; }; for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { if (grid2[i][j]) { ans += dfs(i, j); } } } return ans; } };

class Solution: def countSubIslands(self, grid1: List[List[int]], grid2: List[List[int]]) > int: def dfs(i: int, j: int) > int: ok = grid1[i][j] grid2[i][j] = 0 for a, b in pairwise(dirs): x, y = i + a, j + b if 0 <= x < m and 0 <= y < n and grid2[x][y] and not dfs(x, y): ok = 0 return ok m, n = len(grid1), len(grid1[0]) dirs = (1, 0, 1, 0, 1) return sum(dfs(i, j) for i in range(m) for j in range(n) if grid2[i][j])

func countSubIslands(grid1 [][]int, grid2 [][]int) (ans int) { m, n := len(grid1), len(grid1[0]) dirs := [5]int{1, 0, 1, 0, 1} var dfs func(i, j int) int dfs = func(i, j int) int { ok := grid1[i][j] grid2[i][j] = 0 for k := 0; k < 4; k++ { x, y := i+dirs[k], j+dirs[k+1] if x >= 0 && x < m && y >= 0 && y < n && grid2[x][y] == 1 && dfs(x, y) == 0 { ok = 0 } } return ok } for i := 0; i < m; i++ { for j := 0; j < n; j++ { if grid2[i][j] == 1 { ans += dfs(i, j) } } } return }

function countSubIslands(grid1: number[][], grid2: number[][]): number { const [m, n] = [grid1.length, grid1[0].length]; let ans = 0; const dirs: number[] = [1, 0, 1, 0, 1]; const dfs = (i: number, j: number): number => { let ok = grid1[i][j]; grid2[i][j] = 0; for (let k = 0; k < 4; ++k) { const [x, y] = [i + dirs[k], j + dirs[k + 1]]; if (x >= 0 && x < m && y >= 0 && y < n && grid2[x][y]) { ok &= dfs(x, y); } } return ok; }; for (let i = 0; i < m; ++i) { for (let j = 0; j < n; j++) { if (grid2[i][j]) { ans += dfs(i, j); } } } return ans; }